Samples of plastic product were selected from a batch of material, and the tensile strength in kg was measured for three different time durations, with the following results. UTM UTM i. Time durations Tensile strength 11-12 pm THUY 12-1 pm UT 605 602 375 704 465 287 1-2 pm 680 601 602 478 438 370 512 535 590 501 394 473 Estimate the variability due to time durations of tensile strength and the overall variability UTM of the observations. ii. Is there a significance difference in the tensile strength due to different time durations? Use 1% significance level. UTM UTM
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- A random sample of 150 individuals (males and females) was surveyed, and the individuals were asked to indicate their year incomes. The results of the survey are shown below. Income Category Category 1: $20,000 up to $40,000 Category 2: $40,000 up to $60,000 Category 3: $60,000 up to $80,000 Edit Format Table Test at a = .05 to determine if the yearly income is independent of the gender. (CSLO 1,6,7) 12pt Male 10 35 15 V Paragraph BI U A Female 30 15 45 > 2 T² :The quality control engineer at Palmer Industries is interested in estimating the tensile strength of steel wire based on its outside diameter and the amount of molybdenum in the steel. As an experiment, she selected 25 pieces of wire, measured the outside diameters, and determined the molybdenum content. Then she measured the tensile strength of each piece. The results of the first four are recorded in the table. Tensile Strength Outside Diameter Amount of Molybdenum Place (PSI ) Y (mm) X1 (Units) X2 A 11 0.3 6 B 9 0.2 5 C 16 0.4 8 D 12 0.3 7 Using a statistical software package, the QC engineer determined the multiple regression equation to be Y’=-0.5+20X1+1X2. a) Based on the equation, what is the estimated tensile strength of a steel wire having an outside diameter of .35 mm and 6.4 units of molybdenum? b) Interpret the value of b1 in the equation.b) Civil engineers have developed a 3D model to predict the response of jointed concrete pavements to temperature variations. To check this model, they measured the change in traverse strain on six different occasions and compared this to the modelled values. The table below was obtained. Date Jul 25 Aug 4 Aug 16 - p-value, -58 69 35 Sep 3 -32 Oct 26 -40 Nov 3 -83 - conclusion, Percent Use the Summary output below to answer the following question Summary Output 4.4 (iii) Test whether the differences between the measurements are normally distributed. State your: - hypothesis, Ho Data are normally distributed HA Data are not normally distributed : - the test statistic, 99 - Decision (Reject, Retain), 95 90 80 70 60 50 40 30 20 Change in Strain Note for conclusion: -(if you think there is significant evidence data is normally distributed, enter 1) -(if you think there is significant evidence data is not normally distributed, enter O), Summary Output 4.4: 10 Field Measurement 5 1 -40…
- An environmental group took a sample from a river under study to check on the biochemical oxygen demand (BOD) of the organisms living in the river. The BOD test was conducted over a period of time (in days). The resulting data follow: Time (days): 1 2 4 6 8 10 12 14 18 20 BOD (mg/liter) 0.6 0.7 1.5 1.9 2.1 2.6 2.9 3.7 3.7 3.8 23 16 3.5A certain polymer is used for evacuation systems for aircraft. It is important that the polymer be resistant to the aging process. Twenty specimens of the polymer were used in an experiment. Ten were assigned randomly to be exposed to the accelerated batch aging process that involved exposure to high temperatures for 10 days. Measurements of tensile strength of the specimens were made and the following data were recorded on tensile strength in psi. No aging: 227, 222, 218, 217, 225, 218, 216, 229, 228, 221 Aging: 219, 214, 215, 211, 209, 218, 203, 204, 201, 205 Calculate the sample standard deviation strength of Aging group and no aging groupA study of Spring Peeper breeding yielded the results in the table below. Spring Peepers were bred in different temperature controlled environments. The concentration of dissolved oxygen was measured in each environment. Temperature of water (⁰C) Dissolved oxygen in freshwater (mg/L) Number of eggs hatched into tadpoles 8 10.0 80 8 12.0 100 8 13.5 120 8 16.0 150 15 8.0 600 15 9.5 1090 15 12.0 1700 15 14.0 2190 22 6.5 870 22 8.5 1450 22 10.5 1970 22 12.0 2450 29 5.5 110 29 7.0 210 29 9.0 400 29 11.0 500 Graph the data in the table linked in the description above. (Hint: You must identify the dependent variable and two independent variables.) You may complete your graph electronically or by hand (then take a photo of your graph). Upload your photo or electronic file. (COMPLETE IN ONE GRAPH).
- The "spring-like effect" in a golf club could be determined by measuring the coefficient of restitution (the ratio of the outbound velocity to the inbound velocity of a golf ball fired at the clubhead). Twelve randomly selected drivers produced by two clubmakers are tested and the coefficient of restitution measured. The data follow: Club 1: 0.8406, 0.8104, 0.8234, 0.8198, 0.8235, 0.8562, 0.8123, 0.7976, 0.8184, 0.8265, 0.7773, 0.7871 Club 2: 0.8305, 0.7905, 0.8352, 0.8380, 0.8145, 0.8465, 0.8244, 0.8014, 0.8309, 0.8405, 0.8256, 0.8476 Test the hypothesis that both brands of ball have equal mean overall distance. Use α = 0.05 and assume equal variances. Question: Reject H0 if t0 < ___ or if t0 > ___.The manager of a fleet of automobiles is testing two brands of radial tires. She assigns one tire of each brand at random to the two rear wheels of eight cars and runs the cars until the tires wear out. The data are shown here (in km). Find a 99% CI on the difference in mean life. Which brand would you prefer, based on this calculation? Car Brand 1 Brand 2 1 36925 34318 2 45300 42280 3 36240 35500 4 32100 31950 5 37210 38015 6 48360 47800 7 38200 37810 8 33500 33215A large automobile insurance company selected samples of single and married male policyholders and recorded the number who made an insurance claim over the preceding three-year period. Single Policyholders Married Policyholders ni = 412 n2 = 778 Number making claims = 75 Number making claims = 111 a. Use a= 0.05. Test to determine whether the claim rates differ between single and married male policyholders. z-value (to 2 decimals) p-value (to 4 decimals) We cannot conclude that there is the difference between claim rates. b. Provide a 95% confidence interval (to 4 decimals) for the difference between the proportions for the two populations. Enter negative answer as negative number.